CLAUSIUS-CLAPERYON EQUATION
I. Review
The exact Clapeyron equation, dp / dT  S / V , was obtained for two-phase
equilibrium in a one-component system.
II. Lecture topics
*A. Clapeyron Equation for solid-liquid and solid-solid equilibria
*B. Approximate Clausius-Clapeyron Equation for liquid-gas and solid-gas
d ln p / d (1/ T )  H / R
C. Effect of hydrostatic pressureon vapor pressure
A. Clapeyron Equation for Condensed Phases
By choosing b to be the high-temperature phase, DH is always positive and the slope of
the coexistence line depends on the sign of DV. For melting, most DV = Vliq –Vsol > 0
but in a few cases (H2O, Ga, Ge, Si, Bi) DV < 0. In many cases of solid-solid or
solid-liquid, the coexistence line is approximately linear
B. Vaporization and Sublimation
For X(l) = X(g) [or X(s) = X(g)], one always has DH > 0 and DV > 0, and the
coexistence line is curved. We can rewrite the exact Clapeyron equation in the form
Where Z  Z g  Zl  pV / RT is the difference in compressibility factors for gas and
liquid. Empirically one finds that a plot of ln p vs. 1/T is close to linear. In a range of low
pressures and temperatures (far from the critical point), one can obtain  H from the
slope of ln p vs. 1/T:
If we go one step further and
3. assume DH is independent of T,
it is possible to integrate Eq. (4) to obtain
Trouton's Rule
is an empirical rule that works roughly (to within ±10%) for a wide range of normal
liquids. Tb is the boiling point of the liquid at p = 1 atm. Trouton's rule can be
rationalized on the basis of the ratio of the "free volumes" of the gas (ca. 30,000
3
-1
3
-1
cm mole ) and the liquid (ca. 10% of Vliq, or 3 cm mole ). From statistical mechanics
(5.62),
close to the empirical value !
C. Effect of inert gas pressure on the vapor pressure p of liquid A
At constant temperature T:
po is vapor pressure of A in the
absence of inert gas
P = p + pinert is total pressure if inert
gas present at partial pressure pinert
At equilibrium μA(g,T,p) = μA(l,T,P) and
where the gas phase is treated as a mixture of ideal gases. One can rewrite Eq. (7a) as
and integrate to obtain
This effect is small in practice. For example, po = 0.27 Torr for Hg at 100°C. The vapor
pressure is p = 0.2701 Torr when P = 1 bar and p = 0.283 Torr when P = 100 bar.
D. Sample Problem
–3
Liquid water has a triple point at 6.11 x 10 bar and 273.16K and its normal boiling
point is 1.013 bar (= 1 atm) and 373.15K. The latent heat of vaporization  H vap varies
–1
–1
with temperature, ranging from 45,050 J mol at ~273K to 40,660 J mol at ~373K. The
vapor and liquid densities (and therefore molar volumes) are not available. Predict the
vapor pressure p in bar of liquid water at T = 20°C, 30°C, 80°C, 90°C.
The approximate Clausius-Clapeyron equation can be used, but you should give
the physical assumptions on which it is based.
Hvap / RT 2
Answer: is based on (d ln p / dT )
(a) V vap  V gas  V liq
V gas and (b) V g
RT / P .furthermore, we will use the
integrated form
based on (c)  H vap independent of T over a short range of T.
Which we can compare with actual measured values below:
T2 = 20°C = 293.15K
–3
–3
–3
ln(p/6.11 x 10 ) = –5418.6 (3.411 x 10 – 3.661 x 10 )
= 1.353
–3
p = 3.871 (6.11 x 10 ) = 0.0237 bar at 20°C 0.0233 bar
T2 = 30°C = 303.15K
–3
–3
–3
ln(p/6.11 x 10 ) = –5418.6 (3.299 x 10 – 3.661 x 10 )
= 1.963
–3
p = 7.122 (6.11 x 10 ) = 0.0435 bar at 30°C 0.0424 bar
T2 = 80°C = 353.15K Now use  H vap at 100°C!
–3
–3
ln p = –4890.5 (2.832 x 10 – 2.680 x 10 )
= –0.742 or p = 0.476 bar at 80°C 0.4733 bar
T2 = 90°C = 363.15K
–3
–3
ln p = –4890.5 (2.754 x 10 – 2.680 x 10 )
= –0.360 or p = 0.697 bar at 90°C 0.701 bar
–1
Note: If one assumed  H vap = a + bT, then a = 57,041.5 J and b = –43.90 J K .
Integrating the approximate Clapeyron equation for this choice of  H yields
which is a better integrated form.